# Diff 2^x and 3^x

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i need help when differiation of function in the form a^x.

For example how to i work out the differiational of 2^x and 3^x?

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Use the fact that [imath]a^x = e^{x\log a}[/imath]. This is easy to differentiate.

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I don't get how that helps, the problem of the x being in the exponent is still there.

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The point is that you know (presumably from elementary analysis) what the derivative of ex is.

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Ok, I am new here so I do not know what standard convention is, and if this is an inappropriate response feel free to delete it; but in answer to the original question, to differentiate a^x you can use impicit differentiation. (The other idea posted is probably somewhat easier, but someone seemed not to understand it so this is another way to look at it) That is, if

y = a^x

ln(y) = ln(a^x)

ln(y) = xln(a) (now differentiate impicitly)

y'/y = ln(a)

y' = y ln(a) (substitute the original equation for y)

y' = (a^x)ln(a)

so hopefully that makes sense.

Incidentally, forgive the new guy here, but how are people writing their mathematical stuff so nicely as I do not see the proper tools here in the reply box, and if I try to paste in from somewhere else it does not work?

Edit: I also now notice that this is a very old post. Sorry....

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Incidentally' date=' forgive the new guy here, but how are people writing their mathematical stuff so nicely as I do not see the proper tools here in the reply box, and if I try to paste in from somewhere else it does not work?

[/quote']

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$y=a^x$

$a^x=e^{xlna}$

we can seperate the exponent into

$y=e^xe^{lna}$

and we know

$e^{lna}=a$

so that makes our equation for y

$y=ae^x$

and we know that the derivative of $e^x$ is $e^x$

so that

$\frac{dy}{dx}=ae^x$

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Use the fact that [imath]a^x = e^{x\log a}[/imath']. This is easy to differentiate.

shouldn't that log be ln?

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thanks for catching that.

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$a^x=e^{x{\ln{a}}}$

$\frac{d(e^u)}{dx}={e^u}{\frac{du}{dx}}$

so, $\frac{d(a^x)}{dx}={a^x}{\ln{a}}$

if you felt like it, you could do it logarithmically.

$\ln{y}=x\ln{a}$

$\frac{d(\ln{u})}{dx}={\frac{1}{u}}{\frac{du}{dx}}$

so, $\frac{d(a^x)}{dx}={a^x}{\ln{a}}$

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Don't mind me, I am just trying to write my previous stuff in this LaTex and see if I can make it work.

$y = a^x$

$\ln{y} = \ln{a^x}$

$\ln{y} = x\ln{a}$

now, differentiate implicitly

$\tfrac{dy}{dx}(\tfrac{1}{y}) = \ln{a}$

$\tfrac{dy}{dx} = y\ln{a}$

now, substitute for y

$\tfrac{dy}{dx} = a^x\ln{a}$

so,

$\tfrac{d}{dx}a^x = a^x\ln{a}$

Allright, I think I got it. Thanks!

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shouldn't that log be ln?

Only if you are unfortunate enough to be trapped in a first year calculus class.

"log" denotes the natural logarithm in higher maths. Sometimes it will mean "base-2" if you're trapped in a computer science class (sometimes "lg" is used here). Universal meaning in higher maths: "log" without an explicitly mentioned base means "whatever base makes this statement true" if it matters and "whatever the readers favorite base is" if it doesn't.

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i've always been told log without a specified base is base 10....and my calculator seems to agree.

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I've also been told that at higher level maths people use log to denote ln. Apparently at that level who needs base 10 when you have base e?

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i've always been told log without a specified base is base 10....and my calculator seems to agree.

Now you know better. Go to the library and browse through some advanced math texts and see for yourself (anything beyond elementary calculus). If you like, you can write the authors and let them know your calculator says their notation is wrong. Bring alot of stamps though, you'll need them.

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Yes, but in the context of the original post, which did appear to be in a calculus class, it probably was a good idea to clarify it. I mean, just by the question being asked it was fairly apparent that this was not a discussion in higher math. It was a calculus discussion. So I think it much better to denote $log_e(x)$ by $ln(x)$

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I would write (and expect) $\log$ for base-e in particle physics theory papers.

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they why teach no specified base as base 10 at lower levels? why use ln notation at all?

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Hang over from the old days, laziness, the fact you don't want to explain what e is, easier to comprehend, inaccuracy, the fact that logs to different bases only differ by a constant multiple anyway so it doesn't actually matter arithmetically what one you use (ie the all behave the same arithmetically), the fact that base 10 is what people know and only know at that stage, necessity. Take your pick or invent your own reason.

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how far in maths do you have to go until log goes from log10 to loge?

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Anything beyond intro calculus texts and "ln" has largely vanished, though even in these texts it's not a given that you'll see "ln".

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There also is the whole thing of context. If there is an exponential in the expression, you can pretty much assume that "log" is base e. Even at lower levels. Like in the expression that started this whole thing.

[imath]

a^x = e^{x\log a}

[/imath]

It is pretty obvious what is meant. The clarification is pretty much unnecessary, but why bother confusing a Calc 1 student over it? Incidentally, in my opinion most of the reason that math books don't bother writing "ln" is because in straight math I do not think that you will hardly ever see any log but base e after you get into calculus. In the few astronomy classes that I took though, they used both base 10 (written as log) and base e (written as ln).

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